This question is about the development of potassium-based cells. The value of E⊖ E^\ominus\,E⊖ for potassium suggests that a potassium cell could have a large electromotive force (EMF).
The table below shows some standard electrode potential data:
| Electrode half-reaction | E⊖/VE^\ominus / \text{V}E⊖/V |
|---|---|
| K+(aq)+e−→K(s)\text{K}^+(\text{aq}) + \text{e}^- \rightarrow \text{K}(\text{s})K+(aq)+e−→K(s) | -2.93 |
| 2H2O(l)+2e−→H2(g)+2OH−(aq)2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})2H2O(l)+2e−→H2(g)+2OH−(aq) | -0.83 |
| 12I2(s)+e−→I−(aq)\displaystyle \frac{1}{2}\text{I}_2(\text{s}) + \text{e}^- \rightarrow \text{I}^-(\text{aq})21I2(s)+e−→I−(aq) | +0.54 |
Use data in the table to explain why an aqueous electrolyte is not used for a potassium-based cell.
Potassium-iodine reactant cells have been proposed as energy storage systems. Potassium iodide is the final product of the cell reaction. Use the data in the table to calculate the standard cell EMF of a standard potassium-iodine cell.
An EMF value measured for a non-standard commercial potassium-iodine cell is 2.95 V. Suggest why this value is different from the value calculated in part 2.
In some potassium cells, potassium hexafluoroarsenate (KAsF6\text{KAsF}_6KAsF6) is used as the electrolyte. Deduce the oxidation state of arsenic in KAsF6\text{KAsF}_6KAsF6.
In other potassium-ion cells, potassium manganese oxide (KMnO2\text{KMnO}_2KMnO2) electrodes and potassium electrodes are used. Give an equation for the reaction that occurs at the positive potassium manganese oxide electrode during discharge.
Give an equation for the reaction that occurs at the negative potassium electrode during discharge.
34 exam-style questions on AQA A Level Chemistry Electrode potentials and electrochemical cells (A-level only), covering Electrode potentials and cells (A-level only) and Commercial applications of electrochemical cells (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.